But after placing 27 in (8,8), cornering in (9,8) would lead toa contradiction. So 27 must go into (9,8).

This is easier. There are only 2 possible cells on board for 25: (8,8) & (9,8). Also they are the only 2 possible cells on board for 27. As a result, (8,8) & (9,8) must be {25,27}. (This is analogical to a "hidden pair" in sudoku solving. )

As a result, there is only 1 possible cell on board for 29: (8,9).

And then there is only 1 possible cell on board for 30: (7,8).

Now "gateway" will force (7,9)+(8,8) to be {21..25}, leaving (9,8) as the only possible cell for 27.

Have you (or your solver) tried djape's Jan 20 puzzle? Apparently it requires a new technique, which I call "path shaping". It's hard to describe the technique formally, so here is how I solved it, see if you can get the essence of that technique:

_*: r3c67={_A,_B} (can be considered as a "double gateway")_#: r4c456={_A,_B,_C}_@: r5c3457={_A,_B,_C,_D)=> r5c8=02Also r4c4 (_#) can't touch r3c67 (_*)=> r4c4 can't be from _A or _B, must be from _C=45=> _#: r4c56={_A,_B}

Now r5c7 (_@) must be from _A or _BBut it can't be from _A (too far away from 13 & 19)=> r5c7 (_@) must be from _B=> r4c6 (_#) must be from _B=> r4c5 (_#) must be from _A=> r3c6 (_*) must be from _A=> r3c7 (_*) must be from _B=> _@: r45c345={_A,_C,_D}

Now one of r6c34 (_$) must be from _D=31=> r6c34 (_$) can't be both from _A=> one of r5c34 (_@) must be from _A(otherwise the path for _A will have at least 6 cells, too long)=> r5c4 (_@) must be from _A=> r5c3 (_@) must be from _D=32=> r5c5 (_@) must be from _C=46=> r6c3 (_$) must be from _A=18=> r6c4 (_$) must be from _D=31

Now [04,03] must be @ either r6c67 or r6c78Gateway: r6c7 is reserved for [04,03]Also 54 must be @ either r7c8 or r8c7=> [59..57] can't be @ r67c8+r8c7=> 59 can't be @ r6c8, must be @ r6c6=> [04,03] must be @ r6c78=> 47 must be @ r6c5

I'd like to call these paths (A/B/C/D) fragments.Let's say:Bottleneck means:a given fragment needs a given empty cell as pass-through.What you're pointing out is:n given fragments need n given cells as pass-through.So multiple-bottleneck, allowing to exclude not-involved numbers from the involved cells.

You mentioned one more thing:If a cell is occupied by a fragment, than adjacent cells are occupied by the same fragment if they are necessary pass-throughs.

udosuk wrote:Hope your solver had a easier time than me.

It did; naked singles are quite dominant and not always very elegant in Hidato.

(Hint: I found it easier applying this approach by listing all the unused numbers separately and then crossing out each of them as they're being inserted into the grid. Well things to do when you don't have a program to help. )

Highlight below to see the solution of the puzzle above, I wrote:04|03|50|48|47|46|45|44|3902|05|51|49|54|56|43|40|3806|01|52|53|55|57|42|41|3763|07|08|09|10|58|12|13|3664|62|61|60|59|11|34|35|1467|65|69|26|28|29|33|16|1566|68|70|27|25|80|30|32|1775|73|77|71|79|24|81|31|1874|76|72|78|23|22|21|20|19

I may know a new technique, which is quite obvious for a human player, but I didn’t program it yet. I’d like to call it fuzzy distance.

Do you agree with the logic?

Type I based on straight distance:

For any empty cell (i,j) and for any unplaced number n:Define d as the smallest straight distance from (i,j) to any of the cells where n can be placed.Then for any n2:(n2 > n - d and n2 < n + d) -> then n2 can be excluded from (i,j)

Type I based on shortest-path-distance:

For any empty cell (i,j) and for any unplaced number n:Define d as the smallest shortest-path-distance from (i,j) to any of the cells where n can be placed.Then for any n2:(n2 > n - d and n2 < n + d and n and n2 are in the same fragment) -> then n2 can be excluded from (i,j)

Type II based on straight distance:

For any pair of empte cells (i,j) and (i2,j2)Define n_min and n_max as the lowest resp. the highest number that can be placed in (i,j).Define d as the straight distance between (i,j) and (i2,j2)Then for any n:if n < n_min + d and n > n_max - d then n can be excluded from (i2,j2)

Type II based on shortest-path-distance:

For any pair of empte cells (i,j) and (i2,j2)Define n_min and n_max as the lowest resp. the highest number that can be placed in (i,j).Define d as the shortest-path-distance between (i,j) and (i2,j2)If n_min and n_max are in the same fragment, then for any n in that same fragment:if n < n_min + d and n > n_max - d then n can be excluded from (i2,j2)

Highlight below to see the solution to evert's latest puzzle that I wrote:16|14|13|12|07|80|79|02|0115|17|11|08|81|06|78|03|7620|18|48|10|09|05|04|77|7519|21|49|47|70|45|44|73|7423|22|50|69|46|71|72|43|4224|25|51|67|68|29|39|37|4154|52|26|66|28|30|38|40|3655|53|58|27|65|64|31|33|3556|57|59|60|61|62|63|32|34

I had a great time solving this puzzle, thought about writing a walkthrough but not enough energy.

Could you give 2 examples how you applied each of the "types I & II straight distance" in this puzzle? I think some physical demonstration can be very helpful to explain your "fuzzy logic" instead of the cryptic formulas you wrote above.

udosuk wrote:Could you give 2 examples how you applied each of the "types I & II straight distance" in this puzzle? I think some physical demonstration can be very helpful to explain your "fuzzy logic" instead of the cryptic formulas you wrote above.

An easy example of type I would be:If number 5 can only go in R1C1, R1C2 or R1C3, then 6 could never go in R9C9.

r5c1 must be 29 or 30.As a result, the numbers [27..32] are locked in cells within 3 cells away from r5c1 (27 being 30-3 and 32 being 29+3).Hence r9c1, being 4 cells away from r5c1, can't have [27..32].